Emma's question at Yahoo! Answers regarding solving a trigonometric equation

MarkFL

Staff member
Here is the question:

Solve the following trig equation: 0<=X<=360
cos^2 X -8sinXcosX +3=0

I have posted a link there to this topic so the OP can see my work.

MarkFL

Staff member
Hello Emma,

We are given to solve:

$$\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0$$ where $$\displaystyle 0^{\circ}\le x\le360^{\circ}$$

If we use the following double-angle identities for sine and cosine:

$$\displaystyle \cos(2\theta)=2\cos^2(\theta)-1\,\therefore\,cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$$

$$\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)$$

then the equation becomes:

$$\displaystyle \frac{1+\cos(2x)}{2}-4\sin(2x)+3=0$$

which we can arrange as:

$$\displaystyle 8\sin(2x)-\cos(x)=7$$

Now, if we define (where $$\displaystyle k\in\mathbb{R},\,0^{\circ}<\alpha<90^{\circ}$$):

$$\displaystyle 8=k\cos(\alpha)$$

$$\displaystyle 1=k\sin(\alpha)$$

then we obtain by division:

$$\displaystyle \tan(\alpha)=\frac{1}{8}\,\therefore\,\alpha=\tan^{-1}\left(\frac{1}{8} \right)$$

$$\displaystyle 8^2+1^2=65=k^2\left(cos^2(\alpha)+\sin^2(\alpha) \right)=k^2\,\therefore\,k=\sqrt{65}$$

Hence, our equation becomes:

$$\displaystyle \cos(\alpha)\sin(2x)-\sin(\alpha)\cos(x)=\frac{7}{k}$$

Using the angle-difference identity for sine, and substituting for $\alpha$ and $k$ we obtain:

$$\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right) \right)=\frac{7}{\sqrt{65}}$$

Because of the periodicity of the sine function, we may write, where $$\displaystyle k\in\mathbb{Z}$$

(1) $$\displaystyle \sin\left(2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ} \right)=\frac{7}{\sqrt{65}}$$

Now, combining this with the identity $$\displaystyle \sin\left(180^{\circ}-\theta \right)=\sin(\theta)$$ we also have:

(2) $$\displaystyle \sin\left(\tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ} \right)=\frac{7}{\sqrt{65}}$$

Taking the inverse sine of both sides of (1), we find:

$$\displaystyle 2x-\tan^{-1}\left(\frac{1}{8} \right)+k\cdot360^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)$$

Solving for $x$, we find:

$$\displaystyle x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)-k\cdot180^{\circ}$$

Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:

$$\displaystyle k=0\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right) \approx33.69006752598^{\circ}$$

$$\displaystyle k=-1\implies x=\frac{1}{2}\left(\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)+\tan^{-1}\left(\frac{1}{8} \right) \right)+180^{\circ} \approx213.69006752597983^{\circ}$$

Taking the inverse sine of both sides of (2), we find:

$$\displaystyle \tan^{-1}\left(\frac{1}{8} \right)-2x+(2k+1)\cdot180^{\circ}=\sin^{-1}\left(\frac{7}{\sqrt{65}} \right)$$

Solving for $x$, we find:

$$\displaystyle x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+(2k+1)\cdot90^{\circ}$$

Thus, for appropriate values of $k$, we find the solutions on the given interval for $x$ of:

$$\displaystyle k=0\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+90^{\circ} \approx63.43494882292201^{\circ}$$

$$\displaystyle k=1\implies x=\frac{1}{2}\left(\tan^{-1}\left(\frac{1}{8} \right)-\sin^{-1}\left(\frac{7}{\sqrt{65}} \right) \right)+270^{\circ} \approx243.43494882292202^{\circ}$$

MarkFL

Staff member
While I was busy composing the above posts, someone else at Yahoo! Answers replied with a much simpler method, which I will outline for the benefit of our members.

We are given to solve:

$$\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3=0$$ where $$\displaystyle 0^{\circ}\le x\le360^{\circ}$$

Using a Pythagorean identity, we may write the equation as:

$$\displaystyle \cos^2(x)-8\sin(x)\cos(x)+3\left(\sin^2(x)+\cos^2(x) \right)=0$$

We may arrange this as:

$$\displaystyle 4\cos^2(x)-8\sin(x)\cos(x)+3\sin^2(x)=0$$

Factoring, we obtain:

$$\displaystyle \left(2\cos(x)-\sin(x) \right)\left(2\cos(x)-3\sin(x) \right)=0$$

From the first factor, we obtain the solutions (where $$\displaystyle k\in\mathbb{k}$$):

$$\displaystyle x=\tan^{-1}(2)+k\cdot180^{\circ}$$

and for appropriate values of $k$, we obtain:

$$\displaystyle k=0\implies x=\tan^{-1}(2)\approx63.43494882292201^{\circ}$$

$$\displaystyle k=1\implies x=\tan^{-1}(2)+180^{\circ}\approx243.43494882292202^{\circ}$$

From the second factor, we obtain the solutions:

$$\displaystyle x=\tan^{-1}\left(\frac{2}{3} \right)+k\cdot180^{\circ}$$

and for appropriate values of $k$, we obtain:

$$\displaystyle k=0\implies x=\tan^{-1}\left(\frac{2}{3} \right)\approx33.69006752598^{\circ}$$

$$\displaystyle k=1\implies x=\tan^{-1}\left(\frac{2}{3} \right)+180^{\circ}\approx213.69006752597977^{ \circ}$$

I sure wish I had realized this method first!